I have learned that chirality is a concept, that appears for $(A,B)$ representations of the Lorentz group, where $A\neq B$.

An example would be a Dirac spinor, corresponding to the representation $(\tfrac{1}{2},0)\oplus(0,\tfrac{1}{2})$, where we can identify left- and right-chiral components.

Wikipedia lists the electromagnetic field strength tensor $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$ as transforming under the $(1,0)\oplus(0,1)$ representation of the Lorentz group.

Supposing my first sentence is true, where can I see chirality in the electromagnetic field strength tensor?


3 Answers 3


On can define the dual of a field strength tensor by $$ F^*_{\mu\nu}= \frac 12 \epsilon_{\mu\nu\alpha\beta} F_{\alpha\beta} $$ and one can impose the condition that $F^*_{\mu\nu} =F_{\mu\nu}$ (self dual) or $F^*_{\mu\nu}=-F_{\mu\nu}$. These conditions are preserved by Lorentz tranformations and correspend to the $(1,0)$ and $(0,1)$ representations.

See the table in the Wikipedia article on the "Representation theory of the Lorentz group"

It is possible to write Maxwells' "curl" equations so that they look like a spin one version of the the Weyl equation for chiral fermions: $$ i\hbar \frac{\partial \Psi_\pm}{\partial t}= \pm c(\Sigma\cdot {\bf P}) \Psi_\pm $$ where $\Psi_{\pm}$ are three component Riemann-Silberstein vectors $\Psi_\pm\equiv {\bf E}\pm i{\bf B}c $, ${\bf P}= -i\hbar \nabla$ and the spin-one matrices are $[\Sigma_i]_{jk}==-i\epsilon_{ijk}$. I've always assumed, but have not explicitly checked, that the $\Psi_\pm$ are the two chiral components described by the self-dual and anti-self-dual conditions.

Indeed under electromagnetic duality (and with $c=1$) we have $({\bf E,B})\mapsto ({\bf B,-E})$ so ${\bf E}+i{\bf B}\mapsto ({\bf B}-i{\bf E})=-i({\bf E}+i{\bf B})$ and ${\bf E}-i{\bf B}\mapsto i({\bf E}-i{\bf B})$. Note that $(F^*)^*=-F$, so that in Minkowski signature the eigenvalues of the duality transformation are $\pm i$. This means that real-valued EM fields cannot be self-dual in Lorenzian signature. They can be in Euclidean signature where the Lorentz group is replaced by ${\rm SO}(4)$.

  • 1
    $\begingroup$ Amazing. I wonder if it is possible to move in the opposite direction and write Dirac equation as curl-like operators on linear combinations of the Dirac field $\endgroup$
    – lurscher
    Sep 20, 2018 at 15:11
  • $\begingroup$ Where does the definition of the spin-1 matrices come from? $\endgroup$
    – ersbygre1
    Sep 25, 2018 at 8:16
  • $\begingroup$ @stephan. They are the usual spin-1 matrices, but in the $x,y,z$ vector basis rather than the $|j,m\rangle$ basis. $\endgroup$
    – mike stone
    Sep 25, 2018 at 12:25
  • $\begingroup$ @mikestone. So they are connected to the generators of SO(3)? Or am I mixing something up in my head? I‘ve got another question: Where is the connection between the Riemann-Silberstein vectors and the (anti-)self duality of the em field tensor? And shouldn‘t there be an i in these conditions such that $F**=-F$? $\endgroup$
    – ersbygre1
    Sep 25, 2018 at 12:53
  • 1
    $\begingroup$ @stephan. I was a bit vague in my use of (anti)-self dual in my answer. We have $\star \circ \star=1$ in 4d Euclidean space but $\star \circ \star=-1$ in 4d Minkowski signature. The RS vectors are the Minkowski $\pm i$ eigenvectors of $\star$ in Minkowski, so my understanding is that they are the next-best-thing to self-dual and anti-self dual. And yes you can regard the $\Sigma_i$ as ${\rm SO}(3)$ generators and at the same time the $J=1$, ${\rm SU}(2)$ generators. The $J=1$ rep of ${\rm SU}(2)$ lifts to a rep of ${\rm SO}(3)$. $\endgroup$
    – mike stone
    Sep 25, 2018 at 17:59

There's a great existing answer, I just thought I'd check where the "rotation" comes from.

As you know, the electromagnetic field tensor decomposes under $SO(3)$ into two vectors, $\mathbf{E}$ and $\mathbf{B}$, which are preserved under rotation. In fact, any linear combination of $\mathbf{E}$ and $\mathbf{B}$ are preserved under rotations. Now if we add in the boosts, the specific combinations that are preserved under both rotations and boosts are $$\mathbf{E} = \pm i \mathbf{B}.$$ These correspond to the $(1, 0)$ and $(0, 1)$ irreps; they are called self-dual and anti-self-dual fields.

Here we're working with complex-valued electromagnetic fields, i.e. we have $$\mathbf{E} = \mathbf{E}_0 e^{ik \cdot x}, \quad \mathbf{B} = \pm i \mathbf{E}_0 e^{ik\cdot x}.$$ To get representative real-valued solutions, we may take the real part. For a wave propagating along $\hat{\mathbf{z}}$, guessing $\mathbf{E}_0 \propto (1, \pm i, 0)^T$, we find the self-dual and anti-self-dual fields correspond to light waves with clockwise and counterclockwise circular polarization, a clear manifestation of chirality. You can't boost or rotate a clockwise polarized wave into anything but a clockwise polarized wave.


You can represent the EM field tensor in the same Clifford algebra that's used for the Dirac spinor. The self-dual and anti-self-dual parts are then projected out by $\frac12(1\pm γ^5)$, just like the halves of a Dirac spinor. Maxwell's equations (plus the Lorenz gauge condition) are equivalent to a pair of Dirac-like equations: $\rlap/\partial\rlap{\,/}A = \rlap{\,/}F$ and $\rlap/\partial\rlap{\,/}F = -\rlap/J$.

The Proca equation, which describes a massive, sourceless spin-1 field, is equivalent to exactly the Dirac equation, acting on $ψ = \rlap{\,/}F + im\rlap{\,/}A$ or $$ψ = \begin{pmatrix} -C_z & -C_x + i C_y & (A_t + A_z)im & (A_x - iA_y)im \\ -C_x - i C_y & C_z & (A_x + iA_y)im & (A_t - A_z)im \\ (A_t - A_z)im & (-A_x + iA_y)im & \bar C_z & \bar C_x - i \bar C_y \\ (-A_x - iA_y)im & (A_t + A_z)im & \bar C_x + i \bar C_y & -\bar C_z \end{pmatrix}$$ in the Weyl basis, where $C=E+iB$.

Under coordinate changes, these representations transform using the same matrices as Dirac spinors, but by conjugation instead of left multiplication.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.